Which expression is equivalent to 2 (-3x + 5)?

Can somebody help me with this question?

storchak [24]

Answer:

answer is 6 btw does ~ means congruent

3 0

3 years ago

Read 2 more answers

50 students in 4 buses how many in each bus

12345 [234]

50 divided by 4 is 12.5 so there is about 12 or 13 students on each bus.

3 0

4 years ago

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Use the method of "undetermined coefficients" to find a particular solution of the differential equation. (The solution found ma

Naddika [18.5K]

Answer:

The particular solution of the differential equation

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

Step-by-step explanation:

Given differential equation y''(x) − 10y'(x) + 61y(x) = −3796 cos(5x) + 185e6x

The differential operator form $(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

Rules for finding particular integral in some special cases:-

let f(D)y = $e^{ax}$ then

the particular integral $\frac{1}{f(D)} (e^{ax} ) = \frac{1}{f(a)} (e^{ax} ) if f(a)$ ≠ 0

let f(D)y = cos (ax ) then

the particular integral $\frac{1}{f(D)} (cosax ) = \frac{1}{f(D^2)} (cosax ) =\frac{cosax}{f(-a^2)}$ f(-a^2) ≠ 0

Given problem

$(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

Particular integral:-

$P.I = \frac{1}{f(D)}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + \frac{1}{D^2-10D+61}185e^{6x})$

P.I = $I_{1} +I_{2}$

we will apply above two conditions, we get

$I_{1}$ =

$\frac{1}{D^2-10D+61}( −3796 cos(5x) = \frac{1}{(-25)-10D+61}( −3796 cos(5x) ( since D^2 = - 5^2)$ $= \frac{1}{(36-10D}( −3796 cos(5x) \\= \frac{1}{(36-10D}X\frac{36+10D}{36+10D} ( −3796 cos(5x)$

on simplification we get

= $\frac{1}{(36^2-(10D)^2}36+10D( −3796 cos(5x)$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{1296-100(-25)}$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$

$I_{2}$ =

$\frac{1}{D^2-10D+61}185e^{6x}) = \frac{1}{6^2-10(6)+61}185e^{6x})$

$\frac{1}{37}185e^{6x})$

Now particular solution

P.I = $I_{1} +I_{2}$

P.I = $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

8 0

3 years ago

Question

Evgen [1.6K]

Rotating it around what point? Do you have a picture of your homework?

5 0

4 years ago

According to the manufacturer of a backup UPS device, the normal output voltage is 120 volts. The sample of 40 measured voltage

anyanavicka [17]

Answer:

z = 1.83<1.96

null hypothesis is accepted

The sample is came from a population mean

Step-by-step explanation:

Step :-1

The sample of 40 measured voltage amounts from a unit have a mean of 123.59 volts and a standard deviation of 0.31 volts

given sample size n =40

mean of the sample ×⁻ = 123.59 volts

standard deviation of sample σ = 0.31 volts

Step2:-

Null hypothesis :-

the sample is from a population with a mean equal to 120 volts.

H₀ : μ =120

Alternative hypothesis:-

H₁ : μ ≠120

level of significance:- α =0.05

Step 3:-

The test statistic

$z = \frac{x_{-}-mean }{\frac{S.D}{\sqrt{n} } }$

substitute values and simplification

$z = \frac{123.59-120}{\frac{0.31}{\sqrt{40} } }$

on simplification we get the calculated value

z = 1.83

The tabulated value z =1.96 at 0.05 % level of significance

Conclusion:-

Calculated Z < The tabulated value z =1.96 at 0.05 % level of significance

so the null hypothesis is accepted

The sample is came from a population mean

4 0

3 years ago

Which expression is equivalent to 2 (-3x + 5)?​

Which expression is equivalent to 2 (-3x + 5)?